Calculus II - Tangents with Parametric Equations

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Section 9.2 : Tangents with Parametric Equations

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2. Compute \(\frac{{dy}}{{dx}}\) and \(\frac{{{d^2}y}}{{d{x^2}}}\) for the following set of parametric equations.

\[x = {{\bf{e}}^{ - 7t}} + 2\hspace{0.5in}\,\,y = 6{{\bf{e}}^{2t}} + {{\bf{e}}^{ - 3t}} - 4t\]

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The first thing we’ll need here are the following two derivatives.

\[\frac{{dx}}{{dt}} = - 7{{\bf{e}}^{ - 7t}}\hspace{0.5in}\frac{{dy}}{{dt}} = 12{{\bf{e}}^{2t}} - 3{{\bf{e}}^{ - 3t}} - 4\] Show Step 2

The first derivative is then,

\[\frac{{dy}}{{dx}} = \frac{{\displaystyle \,\,\frac{{dy}}{{dt}}\,\,}}{{\displaystyle \,\,\frac{{dx}}{{dt}}\,\,}} = \frac{{12{{\bf{e}}^{2t}} - 3{{\bf{e}}^{ - 3t}} - 4}}{{ - 7{{\bf{e}}^{ - 7t}}}} = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{{12}}{7}{{\bf{e}}^{9t}} + \frac{3}{7}{{\bf{e}}^{4t}} + \frac{4}{7}{{\bf{e}}^{7t}}}}\] Show Step 3

For the second derivative we’ll now need,

\[\frac{d}{{dt}}\left( {\frac{{dy}}{{dx}}} \right) = \frac{d}{{dt}}\left( { - \frac{{12}}{7}{{\bf{e}}^{9t}} + \frac{3}{7}{{\bf{e}}^{4t}} + \frac{4}{7}{{\bf{e}}^{7t}}} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{{108}}{7}{{\bf{e}}^{9t}} + \frac{{12}}{7}{{\bf{e}}^{4t}} + 4{{\bf{e}}^{7t}}}}\] Show Step 4

The second derivative is then,

\[\frac{{{d^2}y}}{{d{x^2}}} = \frac{{\frac{d}{{dt}}\left( {\frac{{dy}}{{dx}}} \right)}}{{\frac{{dx}}{{dt}}}} = \frac{{ - \frac{{108}}{7}{{\bf{e}}^{9t}} + \frac{{12}}{7}{{\bf{e}}^{4t}} + 4{{\bf{e}}^{7t}}}}{{ - 7{{\bf{e}}^{ - 7t}}}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{108}}{{49}}{{\bf{e}}^{16t}} - \frac{{12}}{{49}}{{\bf{e}}^{11t}} - \frac{4}{7}{{\bf{e}}^{14t}}}}\]